Pelican Parts Forums - Probability Challenge

Pelican Parts Forums (http://forums.pelicanparts.com/)

- Off Topic Discussions (http://forums.pelicanparts.com/off-topic-discussions/)

- - Probability Challenge - Can You Solve It? (http://forums.pelicanparts.com/off-topic-discussions/344435-probability-challenge-can-you-solve.html)

Re: Re: Probability Challenge - Can You Solve It?

Quote:

Originally posted by ikarcuaso
2 doors, 1 correct choice
1 divided by 2 = 1/2 = 50%

Problem solved

Just incorrectly:D

Quote:

Originally posted by Seahawk
But it's so simple.

All I have to do is divine from what I know of you: are you the sort of man who would put the new car behind his own door or his contestants? Now, a clever man would put the new car into his own door, because he would know that only a great fool would reach for what he was given. I am not a great fool, so I can clearly not choose the door in front of you. But you must have known I was not a great fool, you would have counted on it, so I can clearly not choose the door in front of me.

:cool:

INCONCEIVABLE!

Re: Probability Challenge - Can You Solve It?

Quote:

Originally posted by LubeMaster77
Snagged from an engineering site I use from time to time...

'On a game show, a contestant is offered the choice of three doors. A goat is behind two of them, a new car behind the third. The contestant picks one door, but the host opens a different door showing a goat. With two doors left, the host again asks, "Which door?"

To maximize the probability to win the car, should the contestant stay with the first guess, or switch to the other door? Explain the reason behind your answer.'

Is this a trick question???

Re: Re: Re: Probability Challenge - Can You Solve It?

Quote:

Originally posted by ikarcuaso
2 doors, 1 correct choice
1 divided by 2 = 1/2 = 50%

Problem solved

Quote:

Originally posted by lendaddy
Just incorrectly:D

Door selected - 1/3 probability
Door not selected - 2/3 probability

Problem re-solved
:p

Hold on!
"Probability Challenge - Can You Solve It?"

The Answer is...

NO.

OK, I have been thinking about this way to much and I think that the actual answer to the problem is that you can’t tell whether the contestant can increase the probability of picking the car by changing doors without more information.
You see, Amail and I were both correct, but it depends upon circumstances.
Read Amail’s explanation of the deck of cards:

Quote:

Originally posted by Amail
Not true - just because there are two choices doesn't mean they are two equal choices.

Say I have a deck of cards, and ask you to pull out the ace of spades, without looking. You choose your card.

I then look at the remaining 51 cards, keep one to myself and turn the other 50 over to show the ace of spades is not among them. I ask if you'd like to keep your card or take mine instead.

Even though you now have a choice of only two cards, do you think you have a 50% chance that keeping your card is the right call?

He is absolutely correct of course.
I have a 1 in 52 chance of picking the Ace first shot. But if he has gone through and selectively eliminated cards then there is a far greater probability that he will end up with the Ace of Spades. In this instance it will considerably increase my chances of holding the Ace if I exchange cards with Amail.

Now consider an alternative scenario with the same deck – read carefully and identify the obvious difference;

Amail has the same deck of cards, and asks me to pull out the ace of spades, without looking. I choose my card.

Amail then, without looking at them before hand, turns over 50 of the remaining 51 cards and does not reveal the ace. Now he holds one card that was left from the 51, and I hold the one card I originally picked. He then asks if I'd like to keep my card or take his instead.

In this instance there is equal probability of either myself of Amail holding the Ace.

So without knowing whether or not the elimination of the cards, or elimination of the door in the original question, was chance or selection it is impossible to say whether or not the probability of picking the car would increase with changing doors

Two doors left, the choice is now 50:50. You removed one door, so its out of the equation.

Monty knows which door the car is behind. Watch closely, he signals to the curtain - depending upon how hot the women is depends on if she gets the car or not. He car is on a movable stage which he directs will go to which curtain after her final selection is made. If she is hot, she wins and Monty gets a kiss and a hug - now watch clearfully - he also wispers something in her ear - "See me after the show for more details!" Que porn music...

Great. So that means all of us (as guys) are pretty much guaranteed (meaning 100%) of not winning the car.

At that price do you want to win?

So this all pretzeled up logic comes in to play in Deal or No Deal. Ever watch that show? Half the fun for me is trying to guess what the "banker" will decide to offer for the contestant's case based on the cases that aren't yet open. I can't even come close to guessing with any consitency.

Amail,

Its probability of expected returns. You have one briefcase with 1,000,000 and 1 with a dollar, the banker is going to offer $500,000. Or if ten briefcases and ten different dollar amounts its 0.1 times $1 +0.1 times $2+ etc.